i need an example of how to do heat transfer and enthalpy of neutralization

Question

volume of 2.11 M HCl- 99.8

initial temp of Hcl- 24 degrees Celsius

volume of 2.05 M NaOH- 99.2

initial temp of NaOH- 24 degrees Celsius

moles of H+ is 0.211 i think?

moles of OH- is 0.203 i think? i don't know if i calculated this correctly

limiting regent- OH-

heat transfer- ? kJ

empathy of neutralization- ? kJ/mol

Answer 1

The neutralization reaction taking place is HCl + NaOH ==> NaCl + H2O

You will need to know the FINAL temperature of the solution which you didn't provide. I'll guide you through the process, but can't answer it without knowing the final temperature.

moles HCl = 99.8 ml x 1 L/1000 ml x 2.11 mol/L = 0.211 moles HCl (you have this right)

moles NaOH = 99.2 ml x 1 L/1000 ml x 2.05 mol/L = 0.203 moles NaOH (you have this right)

Limiting reagent is NaOH (you have this right)

Now to find the heat transfer you would use the formula q = mC∆T where m = mass of solution; C = specific heat of solution; ∆T = change in temperature of the mixture.

m = 99.8 ml + 99.2 ml = 199 ml = 199 g (assuming a density of 1g/ml)

C = 4.184 J/g/degree (assuming solution has same C as water)

∆T = final temperature - initial temperature = final temp - 24 degrees

Solve for q and that is the heat transferred (Joules) or divide by 1000 to get kJ

Enthalpy of neutralization = heat transferred (k)/0.203 moles NaOH = answer in kJ/mole